1. A random sample of n = 13 observations was selected from a normal population.
ID: 3361921 • Letter: 1
Question
1. A random sample of n = 13 observations was selected from a normal population. The sample mean and variance were x = 3.98 and s2 = 0.3215. Find a 90% confidence interval for the population variance 2. (Round your answers to three decimal places.)
_______to__________
2. Suppose that a response can fall into one of k = 5 categories with probabilities
and that n = 300 responses produced these category counts.
66
A. If you were to test this hypothesis using the chi-square statistic, how many degrees of freedom would the test have?
___________degrees of freedom
B. Find the critical value of 2 that defines the rejection region with = 0.05. (Round your answer to three decimal places.)
^2 0.05=
C. Calculate the observed value of the test statistic. (Round your answer to two decimal places.)
2 =
66
Explanation / Answer
(1-alpha)100% confidence interval for 2={(n-1)s2/chi-sq( 1-alpha/2 ,n-1),(n-1)s2/chi-sq( alpha/2 ,n-1) }
90% confidence interval for 2=(12*0.3215/21.03,12*0.3215/5.23)=(0.184,0.738)
chi-sq(0.1/2,12)=21.03 , chi-sq(1-0.1/2,12)=5.23
(B) there are five classes (k)
A. If you were to test this hypothesis using the chi-square statistic, how many degrees of freedom would the test have?
4 degrees of freedom
there are five classes (k), so degree of freedom=k-1=4
B. Find the critical value of 2 that defines the rejection region with = 0.05. (Round your answer to three decimal places.)
2(0.05, 4)=9.49
C. 2 = 7.63
here p1,p2,p3,p4,p5 is not given , let each have equal probability i.e. 0.2
chi-square=sum(O-E)2/E=7.63
category observed(O) Expected(E) (O-E) (O-E)2/E 1 48 60 -12 2.4 2 63 60 3 0.15 3 73 60 13 2.816667 4 50 60 -10 1.666667 5 66 60 6 0.6 sum= 300 300 0 7.633333Related Questions
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